Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n

Question

Sequence S is defined as S n  = (S n-1  +1) + {1 / (S n-1  +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?

(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850

KarishmaB · Accepted Answer

S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101

S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}

S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101  approx

S_3 = 101+ 1 + 1/101 = 102  approx
.
.
 S_{16} = 115 approx

S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720

The sum will be a little more than 1720.
Answer (c)

MacFauz · Answer

S_1 = 100

S_2 = \frac{101^2 + 1}{101} \approx 101  (Since 1 is negligible when compared to  101^2 )

So, the series is almost an arithmetic progression with a=100, d=1,

We have got to find the sum of "n" terms where "n" is 16.

S_{16} = \frac{16}{2}*(2*100 + (16-1)*1)

= 8*215 = 1720

Answer is C.

muralilawson · Answer

i would say the question is wrong or, atleast, not an exact GMAT type question... why to assume N as an integer .. .it is not specified in the question that n is an integer ....n could be 1.2, 1.2,....etc for n>1 when n is not an integer.... i am thinking in the GMAT prospective...what is the source of this qtn?

mbaiseasy · Answer

S1 =  100

S2 =  100 + (1 + \frac{1}{101})  
  If you will notice  1 + \frac{1}{101}  is approximately 1... S2 = 100 + 1 is approx. ~ 101

S3 =  101 + (1 + \frac{1}{101}) 
  If you wil notice  1 + \frac{1}{101}  is approximately 1... S3 = 101 + 1 approx. ~ 102

Sum = 16 * 100 + 1 + 2 + 3 + ... + 15 =  1600 + \frac{15(15+1)}{2} = 1600 + 120 = 1720

Answer: C

MacFauz · Answer

Since this is a PS question and not a DS question, we are free to make that assumption. "n" is only a subscript indicating the ordinal number of each term and hence can be taken to be integers.

GMATinsight · Answer

Answer: Option C

Please check the solution in attachment

deep1624 · Answer

I did not get this part 16∗100+15∗16/2

which formula have you used is here?

KarishmaB

KarishmaB · Answer

100 + 101 + 102 + .... + 115

= 100 + (100 + 1) + (100 + 2) + (100 + 3) + .... (100 + 15)

=   +

= 100 * 16 +  15*16/2

The highlighted formula is that the sum of first n positive integers is given by n(n+1)/2.
So sum of first 15 positive integers is 15*16/2

Kinshook · Answer

daviesj

Sequence S is defined as S n  = (S n-1  +1) + {1 / (S n-1  +1)} for all n > 1.
If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?

S_1 = 100 
 S_2 = S_1 + 1 + \frac{1}{S_1 + 1} = 101 + \frac{1}{101} = 101.01 > 101

Q = S_1 + S_2 + .... + S_{16} = 100 + 101.01 + 102.01 + .... + 115.01 = \frac{16}{2} (100+115) + .01*15 = 8*215+.15 = 1720.15

IMO C

Akshaynandurkar · Answer

Write down the terms 1 by 1 to see the patterns
S1 = 100
S2 = 101 + 1/101
S3 = 102 + 1/101 + 1/(102+1/101)
S4 = 103 + 1/101 + 1/(102+1/101) + 1/(.....)

here as you can see the first term of each Sn term is significantly greater than each subsequent term and also terms after 1st is either 0 (for S1) or very close to 0 
so there is no significant impact on summation of all terms of the terms after 1st

hence summation = 100+101+102+...115 = 16 x average of 1st and last terms (since this is AP)
summation = 16 * (100+115)/2 = 8 * 215 = 1720

hence the answer is somewhere between 1700 an 1750

findingmyself · Answer

S1=100, S2={100+1}+{1/101}=101+0.009, S3=102+0.009
Sum of S1..S16= 100,101,102,103...115
(Neglecting 0.009)

Sum S1...S16=Finding sum through Arithematic mean method
(115+100)/2= Mean of equally distanced numbers is (first+last)/2

(115+100)/2=(Sum S1...S16)/16

(Sum S1...S16)= 215*8=1720

Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n

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Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n

Prep Toolkit

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